Metamath Proof Explorer

Theorem fin23

Description: Every II-finite set (every chain of subsets has a maximal element) is III-finite (has no denumerable collection of subsets). The proof here is the only one I could find, from p.94 (writeup by Tarski, credited to Kuratowski). Translated into English and modern notation, the proof proceeds as follows (variables renamed for uniqueness):

Suppose for a contradiction that A is a set which is II-finite but not III-finite.

For any countable sequence of distinct subsets T of A , we can form a decreasing sequence of nonempty subsets ( UT ) by taking finite intersections of initial segments of T while skipping over any element of T which would cause the intersection to be empty.

By II-finiteness (as fin2i2 ) this sequence contains its intersection, call it Y ; since by induction every subset in the sequence U is nonempty, the intersection must be nonempty.

Suppose that an element X of T has nonempty intersection with Y . Thus, said element has a nonempty intersection with the corresponding element of U , therefore it was used in the construction of U and all further elements of U are subsets of X , thus X contains the Y . That is, all elements of X either contain Y or are disjoint from it.

Since there are only two cases, there must exist an infinite subset of T which uniformly either contain Y or are disjoint from it. In the former case we can create an infinite set by subtracting Y from each element. In either case, call the result Z ; this is an infinite set of subsets of A , each of which is disjoint from Y and contained in the union of T ; the union of Z is strictly contained in the union of T , because only the latter is a superset of the nonempty set Y .

The preceding four steps may be iterated a countable number of times starting from the assumed denumerable set of subsets to produce a denumerable sequence B of the T sets from each stage. Great caution is required to avoid ax-dc here; in particular an effective version of the pigeonhole principle (for aleph-null pigeons and 2 holes) is required. Since a denumerable set of subsets is assumed to exist, we can conclude _om e. _V without the axiom.

This B sequence is strictly decreasing, thus it has no minimum, contradicting the first assumption. (Contributed by Stefan O'Rear, 2-Nov-2014) (Proof shortened by Mario Carneiro, 17-May-2015)

Ref Expression
Assertion fin23 A FinII A FinIII


Step Hyp Ref Expression
1 isf33lem FinIII = g | a 𝒫 g ω x ω a suc x a x ran a ran a
2 1 fin23lem40 A FinII A FinIII